An ideal gas has a molar heat capacity at constant pressure of `Cp = 2.5 R`. The gas is kept in a closed vessel of volume `0.0083 m^(3)`, at a temperature of `300 K` and a pressure of 1.6 x 10^(6) Nm^(-2). An amount 2.49 x 10^(4) J of heat energy is supplied to the gas. calculate the final temperature and pressure of the gas.

Answer

`675, 3.6 xx10^6`

Explanation

Vessel is closed. Therefore, `Delta W =0` or `Delta Q =Delta U = nC_V DeltaT `
`Delta T = (Delta Q)/(nC_V) = (DeltaQ)/(((pV)/(RT))(C_P -R))= ((Delta Q)(RT))/(pV (C_p -R))`
Substituting the values, we have
`Delta T =((2.49 xx10^4)(300))/((1.6 xx10^6)(0.0083)(3//2)) = 375 " ":." "T_f = Delta T + T = 675 K`
Further at constant volume
`P_2/P_1 = T_2/T_2" ":." "P_2 = T_2/T_1 P_1 = (675/300) (1.6 xx10^6)" , "=3.6 xx10^6 "N/m"^2`